Question

An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is $0^\circ$ and intensity of light emerged from analyser is $\frac{3}{8} I_0$, the angle of rotation of the light by the solution with respect to analyser is ________ degrees.

Hint:

Recall that light intensity becomes $I_0/2$ after the first polarizer. Then use Malus's Law $I = I_{polarised} \cos^2 \theta$, where $\theta$ is the total angle between the plane of polarization and the analyser axis.

Answer 1

Correct Answer: 30

The intensity of light changes as it passes through each component of the optical setup. Let's analyze it step-by-step:

Step 1: When unpolarized light of intensity $I_0$ passes through the first polarizer, its intensity is halved because it only allows one plane of vibration to pass. Let the intensity after the polarizer be $I_1$:
$$I_1 = \frac{I_0}{2}$$

Step 2: This polarized light then enters an optically active solution. The solution rotates the plane of polarization by an angle $\theta$. After passing through the solution, the light intensity remains $I_1 = \frac{I_0}{2}$, but its vibration plane is now shifted by $\theta$ relative to the polarizer's axis.

Step 3: The light finally reaches the analyzer. Since the polarizer and analyzer were initially parallel ($0^\circ$), the angle between the rotated light's polarization plane and the analyzer's transmission axis is $\theta$. According to Malus's Law, the intensity $I$ transmitted through the analyzer is:
$$I = I_{incident} \cos^2 \theta$$
Substituting the values:
$$\frac{3}{8} I_0 = \frac{I_0}{2} \cos^2 \theta$$
$$\frac{3}{8} = \frac{1}{2} \cos^2 \theta$$
$$\cos^2 \theta = \frac{6}{8} = \frac{3}{4}$$
$$\cos \theta = \frac{\sqrt{3}}{2}$$
$$\theta = 30^\circ$$
Thus, the angle of rotation caused by the solution is 30 degrees.